Question: Prove that for the array prev computed by Dijkstra's algorithm, the edges $\left\{u,\mathrm{prep}\left[u\right]\right\}\left(\mathrm{forall}\in v\right)$form a tree.

Dijkstra’s algorithm is a single-source shortest path problem. It is used in both directed and undirected graph with non negative weight.

Dijkstra's method to find shortest path:

• Begin the initialization process at the root node.
• Update the cost of the adjacent nodes in the table by identifying the adjacent nodes.
• From the table, find the node with the lowest cost and repeat step-2 until all nodes have been traversed.

To prove that a graph is a tree, we must show:

The definition of Dijkstra's algorithm demonstrates this. If there are some node pq , then there is a path from q to p if prev[p] Equals q. If prev[m] for some node m q , then there is a route from q to m. As a result, if there is an edge prev[r], for each node r, there is always apath from q to r. As a result, node q is connected to all other nodes.

Contradiction can be used to establish this. Assume that root q,i.e prev[q] , does not have a parent. It is also know that prev[r] is the alone parent of all other nodes r. If node $v_1,v_2,...,v_n$has a cycle, let's assume ${\mathrm{v}}_1=\mathrm{prev}\left[{\mathrm{v}}_2\right],{\mathrm{v}}_2=\mathrm{prev}\left[{\mathrm{v}}_3\right],...,{\mathrm{v}}_{\mathrm{n}}=\mathrm{prev}\left[{\mathrm{v}}_1\right]$. None of nodes are linked to q, which violates the previousstatement. As a result, there is no cycle.

Hence, the edges $\{\mathrm{u},\mathrm{prep}\left[\mathrm{u}\right]\}(\mathrm{forall}\hat{\mathrm{l}}\mathrm{v})$form a tree if all the nodes are connected and there is no cycle.